Question

A test cross of $F_1$ files +a/ +b produced the following offspring
++/ab = 9
ab/ab = 9
+b/ab = 41
a+/ab = 41

What will be distance between linked gene?

• A. 82 cM
• B. 18 cM (cis)
• C. 20 cM
• D. 18 cM (trans)

Answer 1

The correct option is D 18 cM (trans)

Since, recombination frequency is directly proportional to distance between genes, the values are used to locate genes on a chromosome. Here, the parental genotype is +a/+b. Presence of both wild type and mutant alleles of a gene on the same chromosome is known as trans arrangement. Here, the genes are in trans arrangement which makes options B wrong. The hybrid parent produce four gametes: two parental types (+a and +b) and two recombinants (++ and ab) which in turn produce four genotypes upon testcross: two parental types (+a/ab and +b/ab) and two recombinants (++/ab and ab/ab). Total number of recombinants= 9+9=18 which means that genes are 18cM (trans) apart from each other; option D is the correct answer.